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-rw-r--r--inform_employees.cpp44
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diff --git a/inform_employees.cpp b/inform_employees.cpp
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-// 1376. Time Needed to Inform All Employees
-#include "leetcode.h"
-
-/*
- * a company has 'n' employees with a unique id for each employee from 0 to 'n -
- * 1'. the head of the company is the one with the 'head_id'. each employee has
- * one direct manager given the 'manager' array where 'manager[i]' is the direct
- * manager of the i'th employee, 'manager[head_id] = -1'. also it is guaranteed
- * that the subordination relations have a tree structure. the head of the
- * company wants to inform all the company employees of an urgent piece of new.
- * he will inform his direct subordinates, and they will inform their
- * subordinates and so on until all the employees know about the urgent news.
- * the i'th employee needs 'inform_time[i]' minutes to inform all of his direct
- * subordinates (ie. after 'inform_time[i]' minutes, all his direct subordinates
- * can start spreading the news) return the number of minutes needed to inform
- * all the employees about the urget news.
- */
-
-class Solution {
- int dfs(int i, vector<int> &manager, vector<int> &inform_time) {
- if (manager[i] != -1) {
- inform_time[i] += dfs(manager[i], manager, inform_time);
- manager[i] = -1;
- }
- return inform_time[i];
- }
-
-public:
- int numOfMinutes(int n, int head_id, vector<int> &manager,
- vector<int> &inform_time) {
- int ans = 0;
- for (int i = 0; i < n; ++i)
- ans = max(ans, dfs(i, manager, inform_time));
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vector<int> m1 = {-1}, m2 = {2, 2, -1, 2, 2, 2};
- vector<int> it1 = {0}, it2 = {0, 0, 1, 0, 0, 0};
- printf("%d\n", obj.numOfMinutes(1, 0, m1, it1)); // expect: 0
- printf("%d\n", obj.numOfMinutes(6, 2, m2, it2)); // expect: 1
-}